Exercises — Properties — linearity, first - second shift theorems, scaling

WHAT rule: linearity + the standard block $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$. WHY: the $5$ is a constant, so linearity lets it ride outside the transform untouched. $\mathcal{L}\{5e^{4t}\}=5\cdot \frac{1}{s-4}=\frac{5}{s-4}.$ ROC: valid for $s>4$ (where the defining integral converges).

Block: $\mathcal{L}\{\cos 3t\}=\frac{s}{s^2+9}$ (here $b=3$, so $b^2=9$). $\mathcal{L}\{2\cos 3t\}=\frac{2s}{s^2+9}.$ ROC: $s>0$.

WHAT: first shift, $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$, with $a=-2$. WHY that rule: we see $e^{(\text{const})t}$ multiplying an ordinary function — that is the exact trigger for the $s$-shift. Start: $\mathcal{L}\{\sin 5t\}=\frac{5}{s^2+25}=F(s)$. Replace $s\to s-(-2)=s+2$ everywhere: $\mathcal{L}\{e^{-2t}\sin 5t\}=F(s+2)=\frac{5}{(s+2{)}^2+25}.$ ROC: the shift drags the convergence region with it, so $s>-2$.

Linearity splits the two terms; each uses a standard block. $\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4}$, and $\mathcal{L}\{1\}=\frac{1}{s}$. $\mathcal{L}\{4t^3-7\}=4\cdot \frac{6}{s^4}-7\cdot \frac{1}{s}=\frac{24}{s^4}-\frac{7}{s}.$ ROC: $s>0$ (both blocks need it).

WHAT: second shift, $\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$, here $a=3$. WHY: the argument is already $(t-3)$ and the step is $u(t-3)$ — matched form, so we plug straight in. Underlying $f(t)=t^2\Rightarrow F(s)=\frac{2}{s^3}$. $\mathcal{L}\{(t-3{)}^{2}u(t-3)\}=e^{-3s}\cdot \frac{2}{s^3}.$ ROC: $s>0$ (the $e^{-3s}$ tag does not change convergence).

WHAT: scaling, $\mathcal{L}\{f(ct)\}=\frac{1}{c}F(s/c)$ for $c>0$. Here $f(t)=\cos t$, $c=4$. WHY the $1/c$: the substitution $\tau =4t$ carries a Jacobian $dt=d\tau /4$, which survives as a factor outside. $\mathcal{L}\{\cos 4t\}=\frac{1}{4}F\left(\frac{s}{4}\right)=\frac{1}{4}\cdot \frac{s/4}{(s/4{)}^2+1}.$ Multiply top and bottom by $16$: $=\frac{1}{4}\cdot \frac{4s}{s^2+16}=\frac{s}{s^2+16}.$ ✔ (matches the direct block). ROC: $s>0$.

Rewrite $t^2$ in powers of $(t-3)$ so the second shift applies: $t^2=((t-3)+3{)}^2=(t-3{)}^2+6(t-3)+9.$ Each term now has a genuine $(t-3)$ argument (the constant $9$ counts as $9\cdot 1$, a "$(t-3{)}^0$" piece). With $u(t-3)$ multiplying all, apply $e^{-as}F(s)$, $a=3$, to each block:

• $(t-3{)}^2\to \frac{2}{s^3}$
• $6(t-3)\to 6\cdot \frac{1}{s^2}$
• $9\to 9\cdot \frac{1}{s}$ $\mathcal{L}\{t^{2}u(t-3)\}=e^{-3s}\left(\frac{2}{s^3}+\frac{6}{s^2}+\frac{9}{s}\right).$ ROC: $s>0$.

Step 1 — kill the square (WHY): we have no table entry for ${\cos }^2$, so use the identity ${\cos }^{2}t=\frac{1}{2}(1+\cos 2t)$ to turn it into things we DO know. $\mathcal{L}\{{\cos }^{2}t\}=\frac{1}{2}\mathcal{L}\{1\}+\frac{1}{2}\mathcal{L}\{\cos 2t\}=\frac{1}{2}\cdot \frac{1}{s}+\frac{1}{2}\cdot \frac{s}{s^2+4}.$ Call this $F(s)=\frac{1}{2s}+\frac{s}{2(s^2+4)}$ (ROC $s>0$). Step 2 — apply first shift with $a=3$ ($s\to s-3$): $\mathcal{L}\{e^{3t}{\cos }^{2}t\}=\frac{1}{2(s-3)}+\frac{s-3}{2((s-3{)}^2+4)}.$ ROC: $s>3$ (the shift moves the region right by $3$).

(a) First shift: $f(t)=t\Rightarrow F(s)=\frac{1}{s^2}$. With $a=2$, $s\to s-2$: $\mathcal{L}\{t{e}^{2t}\}=\frac{1}{(s-2{)}^2}.$ (b) Direct table entry $\mathcal{L}\{t^n{e}^{at}\}=\frac{n!}{(s-a{)}^{n+1}}$ with $n=1,a=2$: $\frac{1!}{(s-2{)}^2}=\frac{1}{(s-2{)}^2}$. ✔ Same. ROC: $s>2$.

Two rules stack here. Treat $e^{-t}(t-1{)}^{3}u(t-1)$ as $e^{-t}\cdot h(t)$ where $h(t)=(t-1{)}^{3}u(t-1)$. Step 1 — transform $h$ (second shift, $a=1$, base $f=t^3$, $F=\frac{6}{s^4}$): $H(s)=\mathcal{L}\{h(t)\}=e^{-s}\cdot \frac{6}{s^4}.$ Step 2 — apply first shift for the $e^{-t}$ (so $a=-1$, i.e. $s\to s+1$ in $H$): $\mathcal{L}\{e^{-t}h(t)\}=H(s+1)=e^{-(s+1)}\cdot \frac{6}{(s+1{)}^4}=\frac{6e^{-1}{e}^{-s}}{(s+1{)}^4}.$ Note the $e^{-1}$ (a constant) that pops out when $s\to s+1$ hits the $e^{-s}$ factor — easy to miss. ROC: $s>-1$.

This stacks second shift (the delayed $u(t-\frac{\pi }{6})$ with a matched $(t-\frac{\pi }{6})$ argument) and first shift (the $e^{2t}$ multiplier). Set $a=\frac{\pi }{6}$ consistently throughout. Step 1 — peel off $e^{2t}$. Write the whole thing as $e^{2t}\cdot h(t)$ where $h(t)=\sin 3\left(t-\frac{\pi }{6}\right)u\left(t-\frac{\pi }{6}\right).$ Step 2 — transform $h$ (second shift, $a=\frac{\pi }{6}$). The argument is already $(t-\frac{\pi }{6})$, so no rewriting is needed. Base $f(t)=\sin 3t$, $F(s)=\frac{3}{s^2+9}$: $H(s)=e^{-\pi s/6}\cdot \frac{3}{s^2+9}.$ Step 3 — apply first shift for $e^{2t}$ ($a=2$, so $s\to s-2$ everywhere in $H$): $\mathcal{L}\{e^{2t}h(t)\}=H(s-2)=e^{-\pi (s-2)/6}\cdot \frac{3}{(s-2{)}^2+9}=\frac{3e^{\pi /3}{e}^{-\pi s/6}}{(s-2{)}^2+9}.$ Watch the constant $e^{\pi /3}$ that pops out when $s\to s-2$ hits the $e^{-\pi s/6}$ factor (since $e^{-\pi (s-2)/6}=e^{\pi /3}{e}^{-\pi s/6}$). ROC: $s>2$.

Two tools: first shift for $e^{-t}$, and the $t\times (\cdots )$ multiplier which is $-\frac{d}{ds}$ of the transform (from Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n): $\mathcal{L}\{tg(t)\}=-G^{\prime }(s)$). Step 1 — first shift on $\cos t$ for the $e^{-t}$ ($a=-1$, $s\to s+1$): $\mathcal{L}\{e^{-t}\cos t\}=\frac{s+1}{(s+1{)}^2+1}=:G(s)$. Step 2 — multiply by $t$ ⇒ $-G^{\prime }(s)$. Let $D=(s+1{)}^2+1=s^2+2s+2$, numerator $N=s+1$. $G^{\prime }(s)=\frac{N^{\prime }D-N{D}^{\prime }}{D^2}=\frac{(1)(s^2+2s+2)-(s+1)(2s+2)}{(s^2+2s+2{)}^2}.$ Numerator $=s^2+2s+2-(2s^2+4s+2)=-s^2-2s=-(s^2+2s)$. $-G^{\prime }(s)=\frac{s^2+2s}{(s^2+2s+2{)}^2}.$ So $\mathcal{L}\{t{e}^{-t}\cos t\}=\frac{s^2+2s}{((s+1{)}^2+1{)}^2}$. ROC: $s>-1$.

Step 1 — transform the ODE (using $\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)$ from Solving ODEs with Laplace Transforms, and the second shift on the right, $\mathcal{L}\{u(t-1)\}=\frac{e^{-s}}{s}$): $sY-0+2Y=\frac{e^{-s}}{s}\Rightarrow (s+2)Y=\frac{e^{-s}}{s}\Rightarrow Y(s)=\frac{e^{-s}}{s(s+2)}.$ Step 2 — partial fractions on $\frac{1}{s(s+2)}=\frac{1/2}{s}-\frac{1/2}{s+2}$ (split into table-ready pieces; see Inverse Laplace Transform and Partial Fractions): $Y(s)=e^{-s}\left(\frac{1/2}{s}-\frac{1/2}{s+2}\right).$ Step 3 — invert with the second shift backwards. The $e^{-s}$ means "delay by $1$ and switch on with $u(t-1)$." The bracket inverts to $\frac{1}{2}-\frac{1}{2}{e}^{-2t}$, so evaluate that at $t-1$: $y(t)=\left[\frac{1}{2}-\frac{1}{2}{e}^{-2(t-1)}\right]u(t-1)=\frac{1}{2}(1-e^{-2(t-1)})u(t-1).$ Sanity — all cases: for $t<1$, $u=0$ so $y=0$ (nothing has been switched on yet). At $t=1$, $y=\frac{1}{2}(1-1)=0$ (continuous start). As $t\to \infty$, $y\to \frac{1}{2}$ (steady state where $y^{\prime }\to 0$ gives $2y=1$). ✔

Step 1 — base transform: $f(t)=t{e}^{-t}\Rightarrow F(s)=\frac{1}{(s+1{)}^2}$ (first shift on $\mathcal{L}\{t\}=\frac{1}{s^2}$). Step 2 — scaling, $c=2$: $\mathcal{L}\{f(2t)\}=\frac{1}{2}F(s/2)=\frac{1}{2}\cdot \frac{1}{(s/2+1{)}^2}$. Simplify: ${\left(\frac{s}{2}+1\right)}^2=\frac{(s+2{)}^2}{4}$, so $\mathcal{L}\{f(2t)\}=\frac{1}{2}\cdot \frac{4}{(s+2{)}^2}=\frac{2}{(s+2{)}^2}.$ Verify directly: $f(2t)=2t{e}^{-2t}$, and $\mathcal{L}\{2t{e}^{-2t}\}=2\cdot \frac{1}{(s+2{)}^2}=\frac{2}{(s+2{)}^2}.$ ✔ ROC: $s>-2$.

Step 1 — strip the $e^{-2s}$ (second shift, $a=2$): whatever the rest inverts to, call it $g(t)$; the full answer is $g(t-2)u(t-2)$. Step 2 — invert $\frac{s-1}{(s-1{)}^2+9}$ (first shift, $s\to s-1$ means multiply by $e^t$): the un-shifted $\frac{s}{s^2+9}$ inverts to $\cos 3t$, so with the $-1$ shift: $g(t)=e^t\cos 3t$. Step 3 — combine: ${\mathcal{L}}^{-1}\{\cdots \}=e^{t-2}\cos (3(t-2))u(t-2).$ Cases: for $t<2$ this is $0$; for $t\ge 2$ it is a growing cosine that "starts" at $t=2$. ✔